# 原题

In a row of trees, the `i`-th tree produces fruit with type `tree[i]`.

You start at any tree of your choice, then repeatedly perform the following steps:

2. Move to the next tree to the right of the current tree.  If there is no tree to the right, stop.

Note that you do not have any choice after the initial choice of starting tree: you must perform step 1, then step 2, then back to step 1, then step 2, and so on until you stop.

You have two baskets, and each basket can carry any quantity of fruit, but you want each basket to only carry one type of fruit each.

What is the total amount of fruit you can collect with this procedure?

Example 1:

```Input: [1,2,1]
Output: 3
Explanation: We can collect [1,2,1].
```

Example 2:

```Input: [0,1,2,2]
Output: 3
Explanation: We can collect [1,2,2].
If we started at the first tree, we would only collect [0, 1].
```

Example 3:

```Input: [1,2,3,2,2]
Output: 4
Explanation: We can collect [2,3,2,2].
If we started at the first tree, we would only collect [1, 2].
```

Example 4:

```Input: [3,3,3,1,2,1,1,2,3,3,4]
Output: 5
Explanation: We can collect [1,2,1,1,2].
If we started at the first tree or the eighth tree, we would only collect 4 fruits.
```

Note:

1. `1 <= tree.length <= 40000`
2. `0 <= tree[i] < tree.length`

# 题解

```
map m;

m.insert(make_pair(1, 1)); // 第一个元素代表类型为1，第二个元素代表类型为1的出现的个数

```

```
m.size(); // 当前不同类型的总数

m.find(1)->second; // 类型为1的出现的次数

```

# 代码

```class Solution {
public:
int totalFruit(vector<int>& tree) {
map<int, int> m;
map<int, int>::iterator it;
int l = 0, r = 0, len = tree.size(), a = 0;
while (r < len) {
if (m.size() <= 2) {
it = m.find(tree[r]);
if (it != m.end()) {
it->second++;
m.insert(make_pair(tree[r], it->second + 1));
} else {
m.insert(make_pair(tree[r], 1));
}
r++;
if (a < r - l) a = r - l;
} else {
it = m.find(tree[l]);
m.erase(tree[l]);
if (it->second != 1) {
m.erase(tree[l]);
m.insert(make_pair(tree[l], it->second - 1));
it = m.find(tree[l]);
}
l++;
}
}
return a;
}
};
```